Mathematical Field Notes

Lines through four lines

with 8 comments

(Reposting this from an earlier blog which I gave up on, but liked the post so I added some pictures – all images produced using the amazing free alternative to Maple/Mathematica, Sage).

I was reading Fulton-Pandharipande (“Notes on stable maps and quantum cohomology”) the other day and came across the classical result that there are exactly two lines passing through a generic quadruple of lines in \mathbb{CP}^3. I encourage people to whom this fact is unfamiliar to convince themselves of it. It was unfamiliar to me and I found it hard to visualise, so I sat down and drew some pictures until I understood it.

Suppose for starters that three of the lines (call them A, B and C) lie on a quadric surface. Recall that a quadric surface is ruled in two different ways by families of lines – call these lines alpha- and beta- curves. In the picture on the left they’re yellow and red respectively.

Diagrams illustrating the there are two lines through four lines.

The alpha-curves are all mutually disjoint, the beta-curves are all mutually disjoint and any alpha-curve intersects any beta-curve in exactly one point transversely. Lines in general position in \mathbf{CP}^3 are disjoint, so our three lines are either all alpha- or all beta-curves. Let’s say they’re all alpha-curves for definiteness’s sake.

Our fourth line (D, blue in the picture on the right) intersects the quadric surface in precisely two points. But each intersection point has a unique beta-curve passing through it. These two (red) beta-curves intersect A, B, C (the yellow lines) and D (the blue line) and these are the two lines in question. QED

Now how do we see that any three general lines lie on a quadric surface? Fix a point on line A and draw all the lines from this point which hit line B. Precisely one of these will hit line C. This means that for each point p of A there is a unique line L_p through that point which hits both B and C. These lines trace out a quadric surface containing the three lines as you vary the point on A. Note that the L_p are precisely the beta-curves of the resulting surface.



Written by Jonny Evans

August 14, 2012 at 12:36 pm

8 Responses

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  1. Hello Jonny,

    I too have been reading a lot on Gromov Witten Invariants and trying to gain a little more grip at it.
    And I has this fact only through GW invariants. Thanks for the post for more intuitive explanations with drawings. May I ask you drew these pictures in Sage?


    Vijay Sharma

    September 10, 2012 at 10:06 am

    • Hi Vijay,

      You’re welcome! I think the code I used for the second picture was this (see below, though for some reason Sage is running buggily on my work computer at the moment, so I can’t verify that it works this morning…) – the code is unnecessarily complicated/full of redundancies because I was simultaneously working out on paper the equations for the red lines, learning how to use Sage, and writing the code (very bad practice…).


      p = implicit_plot3d(x^2+y^2-z^2==1, (x, -2, 2), (y, -2,2), (z, -2,2),opacity=0.8,color=’grey’)
      u = var(‘u’)
      q = parametric_plot3d((u, 1, u), (u, -2, 2),color=’yellow’,thickness=5)
      r = parametric_plot3d((u/sqrt(2), u/sqrt(2)+sqrt(2), u+1), (u, -3, 1),color=’yellow’,thickness=5)
      s = parametric_plot3d((u/sqrt(10), -3*u/sqrt(10)+sqrt(10), u-3), (u, 0, 5),color=’yellow’,thickness=5)
      t = parametric_plot3d((u, 0.7, 0.5*u+0.1),(u, -2, 2),color=’blue’,thickness=5)
      n = (0.1-sqrt(0.1^2-4*(1-0.5^2)*(0.7^2-0.1^2-1)))/(2*(1-0.5^2))
      m = 0.5*n+0.1
      k = (0.1+sqrt(0.1^2-4*(1-0.5^2)*(0.7^2-0.1^2-1)))/(2*(1-0.5^2))
      f = 2
      g = -5
      a = parametric_plot3d((-u/sqrt(1+f^2), -f*u/sqrt(1+f^2)+sqrt(1+f^2), u-f), (u, -1, 4),color=’red’,thickness=5)
      b = parametric_plot3d((-u/sqrt(1+g^2), -g*u/sqrt(1+g^2)+sqrt(1+g^2), u-g), (u, -7, -3),color=’red’,thickness=5)
      c = parametric_plot3d((n,0.7,m),(u,-2,2),color=’red’,thickness=15)
      d = parametric_plot3d((k,0.7,0.5*k+0.1),(u,-2,2),color=’red’,thickness=15)


      Let me know if you produce any nice pictures of your own!


      Jonny Evans

      September 10, 2012 at 11:13 am

  2. […] A useful piece of software I encountered was GeoGebra (for producing geometric diagrams, etc.). For some beautiful examples of output (including applets for playing with linkages), see this page, maintained by Chris Sangwin, who introduced us to the software. While it’s true that the same results (and many more) could be obtained with Maple or Sage, it’s also true that GeoGebra is free and intuitive, and it’s extremely quick to produce diagrams of the sort it’s good at (not as code-heavy as Maple). I’m told there’s a 3d version and this would have been useful for producing these diagrams. […]

  3. Jonny,

    Greetings and hope all is well at your side.
    Really thankful for your post and comments to help me to try things.

    It was really useful.

    Jonny after reading this post I felt that there is some connection between knot theory and Gromov Witten Invariants. I will try to summarize my thought process in the next post.

    Thanking you



    October 20, 2012 at 10:19 pm

  4. Jonny,

    Let me know if my earlier post reached you.

    Thanks & Regards,

    Vijay Sharma

    October 23, 2012 at 7:32 pm

  5. “These lines trace out a quadric surface containing the three lines as you vary the point on A” Why a quadric?


    March 22, 2015 at 2:14 pm

    • I think I’m right in saying that PGL(4,C) acts transitively on generic triples of lines in P^3: at least this is heuristically justified by a dimension count – the group has complex dimension 15 and the space of triples of lines has dimension 12. This could be made rigorous if you check that the infinitesimal action is surjective at your favourite configuration of lines (provided your favourite triple is generic!) because it would give you a Zariski open PGL(4,C)-orbit in the space of line-triples (which you now take as your definition of “generic”). It also acts transitively on smooth quadric surfaces (because any two nondegenerate quadratic forms are equivalent over C). Therefore it suffices to fix your favourite configuration of three lines and check that it’s contained in a quadric surface.

      Jonny Evans

      March 24, 2015 at 8:56 am

      • Thanks a lot for the answer, Jonny.


        March 25, 2015 at 1:08 pm

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